b/ \(a-\frac{1}{a}=\sqrt{a}+\frac{1}{\sqrt{a}}\)

\(\Leftrightarrow\sqrt{a}-\frac{1}{\sqrt{a}}=1\)

\(\Leftrightarrow a+\frac{1}{a}-2=1\)

\(\Leftrightarrow a+\frac{1}{a}=3\)

\(\Leftrightarrow a^2+\frac{1}{a^2}+2=9\)

\(\Leftrightarrow\left(a-\frac{1}{a}\right)^2=5\)

\(\Leftrightarrow a-\frac{1}{a}=\sqrt{5}\)

a/ Ta có: \(x=\frac{1-5y}{2}\) thê vô ta được

\(x^2+y^2=y^2+\left(\frac{1-5y}{2}\right)^2=\frac{29y^2-10y+1}{4}\)

\(=\frac{1}{116}\left(29^2y^2-290y+29\right)=\frac{1}{116}\left[\left(29^2y^2-2.29y.5+25\right)+4\right]\)

\(=\frac{1}{116}\left[\left(29y-5\right)^2+4\right]\ge\frac{4}{116}=\frac{1}{29}\)

xét VT = \(\frac{\sqrt{a}-\sqrt{a+1}}{a-a-1}\)   + \(\frac{\sqrt{a+1}-\sqrt{a+2}}{a+1-a+2}\) + \(\frac{\sqrt{a+2}-\sqrt{a+3}}{a+2-a-3}\) 

         =  \(-\)\(\sqrt{a}+\sqrt{a+1}-\sqrt{a+1}+\sqrt{a+2}-\sqrt{a+2}+\sqrt{a+3}\) 

         =   \(\sqrt{a+3}-\sqrt{a}\)

          =   \(\frac{\sqrt{a+3}^2-\sqrt{a}^2}{\sqrt{a+3}+\sqrt{a}}\)

         =\(\frac{a+3-a}{\sqrt{a+3}+\sqrt{a}}\) =\(\frac{3}{\sqrt{a+3}\sqrt{a}}\) = VP \(\Rightarrow\) đpcm

\(\frac{1}{a}+\frac{1}{b}+\frac{1}{c}=0\Rightarrow ab+bc+ca=0\Rightarrow\left(a+c\right)\left(b+c\right)=c^2\)

Vì \(a,b>0\)mà \(\frac{1}{c}=-\left(\frac{1}{a}+\frac{1}{b}\right)< 0\)nên \(c< 0\Rightarrow\sqrt{\left(a+c\right)\left(b+c\right)}=-c\)

\(\Rightarrow2c+2\sqrt{\left(a+c\right)\left(b+c\right)}=0\Rightarrow\left(a+c\right)+2\sqrt{\left(a+c\right)\left(b+c\right)}+\left(b+c\right)=a+b\)

\(\Rightarrow\left(\sqrt{a+c}+\sqrt{b+c}\right)^2=a+b\)---> 2 vế đều dương nên ta lấy căn 2 vế:

\(\sqrt{a+c}+\sqrt{b+c}=\sqrt{a+b}\)

k có số dương nào để tổng trên bằng 0

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